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In the material I'm going over there's a "standard fact" about sequences: $$a_n>0, \lim\limits_{n \to \infty}{a_{n+1} \over a_n} = q \implies \sqrt[n]{a_n}=q$$ Now, once we consider series, the d'Alambert ratio test states: $a_n>0, {a_{n+1} \over a_n} \to q < 1 \implies \sum a_n$ converges.

The Cauchy criterion postulates: $\sqrt[n]{a_n} \to q \implies \sum a_n$ converges.

Now, I've come across a statement that the satisfaction of the Cauchy criterion Cauchy is stronger than d'Alambert. The statement I've seen about Cauchy being stronger that d'Alambert was $$\sqrt[n]{a_n}\to q \implies {a_{n+1} \over a_n} \to q \qquad (1)$$

Two questions here:

  1. Is the original "standard fact" not only an implication but actually an equivalence?
  2. Is $(1)$ plain wrong? If we assume $(1)$ is correct and take $\sqrt[n]{a_n} \to 2$, the series converges by Cauchy but diverges by the ratio test. How do we reconcile this?

I've seen several expositions comparing the root and ratio test so far. None of them has clarified the matter.

Zelazny
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  • The relation between these limits is described e.g. here: http://math.stackexchange.com/a/1164162/259493 – Dominik Nov 30 '16 at 16:11
  • @NickPeterson It seems the notes I'm reading are lacking vital information. Just to check if I got this right, even with the knowledge that the root test implies convergence if $q<1$, we only can write ${a_{n+1} \over a_n} \to q \implies \sqrt[n]{a_n}$ and this means that if the ratio test is satisfied, then the root test is as well but there are cases when the root test is satisfied but the ratio test is not (so the root test is stronger in identifying convergence)? EDIT: Your answer posted in the interim replies to my question :) – Zelazny Nov 30 '16 at 16:13

1 Answers1

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The statement (1) is not correct.

For instance, consider $$ a_n=\frac{1}{2^{n+(-1)^n}}. $$ The ratios are $$ \frac{a_{n+1}}{a_n}=\frac{2^{n+(-1)^n}}{2^{(n+1)+(-1)^{n+1}}}=\begin{cases}2 & \text{$n$ even}\\\frac{1}{8} & \text{$n$ odd}\end{cases} $$ while the roots satisfy $\sqrt[n]{a_n}\to\frac{1}{2}$ as $n\to\infty$.

In fact, the opposite of (1) is the truth: if $\lvert\frac{a_{n+1}}{a_n}\rvert\to q$ as $n\to\infty$, then $\sqrt[n]{\lvert a_n\rvert}\to q$ as $n\to\infty$ as well.

As for your claims about convergence: remember that convergence of the ratios/roots is not sufficient to show convergence in either case; what's required is that the limit be less than 1.

Nick Peterson
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